Why is $\phi(x^i)=y^i$ not a group homomorphism between the cyclic group of order $36$ to the cyclic group of order $17$? In Artin Algebra 2.10.3 there gives a group homomorphism from a cyclic group of order $12$ to a cyclic group of order $6$. Defined by $\phi(x^i)=y^i$ with $x$ in the cyclic group of order $12$ and $y$ in the cyclic group of order $6$. But we notice that for two cyclic groups with their order coprimes (for instance $36$ and $17$). This seems not to be a group homomorphism because we know that the only homomorphism between groups of coprime orders is the trivial homomorphism. 
Could someone tell me why it is not a group homomorphism with out us Lagrange's theorem, but just use the definition of homomorphism. Which condition is this map does not satisfy?
Thank for any help!
 A: In general this will not even define a function between the groups, because the condition will require $\phi$ to take on different values at the same input.
(Notice, for instance, that if $G$ is cyclic of order 37 then there are many different ways of writing the identity: 
$$e = x^0 = x^{37} = x^{74} = x^{-37} = \cdots$$
Do these all give the same condition in general?)
A: Taking directly your example
$$\phi:C_{36}=\langle x\rangle\to C_{17}=\langle y\rangle\;,\;\;\phi(x^k)=y^k$$
For example we'd get
$$1=\phi(x^{36})=\phi(x^{17}x^{19})=\phi(x^{17})\phi(x^{19})=y^{17}y^{19}=1\cdot y^{19}=y^2\implies y^2=1$$
which of course is absurd.
A: Consider $Z_p$ and $Z_q$ where $p$ and $q$ are coprime.
As $Z_p$ and $Z_q$ are cyclic, we have that they each have some generator.  Let's call them $x_p$ and $x_q$.  We have that $x_p^p = e$, and $x_q^q=e$ (these are technically identities on different groups, but this won't be important).
Recall that any group homomorphism $\phi:G\to H$ has $\phi(e)=e$.  Let $\phi:Z_p\to Z_q$ be a group homomorphism.  Then, we have that:
$$\phi(e)=e\implies \phi(x_p^p) = x_q^q\implies \phi(x_p)^p = x_q^q$$
We know that $\phi(x_p)\in Z_q$, and any element of $Z_q$ can be written as $x_q^m$ for some $m\in\mathbb Z/(q-1)\mathbb Z$.  So, we have that $$x_q^q = x_q^{mp} = e$$
Here, we see that $mp = kq$ for some $k$.  As $p$ and $q$ are coprime, it follows that $m = kq$.  So, $\phi(x_p) = x_q^{kq} = e^k = e$, so $\phi$ is the trivial homomorphism.
